Math Revolution Approach (PS)

If 11!/(11-r)!<1,000, what is the greatest possible value of r?

A. 1
B. 2
C. 3
D. 4
E. 5

=> 11! / (11-r)! = 11*(11-1)*…*(11-r+1)

r = 1 : 11 < 1,000
r = 2 : 11*10 = 110 < 1,000
r = 3 : 11*10*9 = 990 < 1,000
r = 4 : 11*10*9*8 = 7290 > 1,000

r = 3

Ans: C

Which of the following is the closest to 11*10^{20}–9*10^{10}?

A. 10^2 B. 10^7 C. 10^{10} D. 10^{20} E. 10^{21}

=>9*10^{10} is a relatively small number compared with 11*10^{20}.
11*10^{20}–9*10^{10} is approximate to 11*10^{20}, which is similar to 10^{21}.

Ans: E

Which of the following inequalities is equal to |x-3|x||<8?

A. 0<x<4
B. 0<x<2
C. -2<x<4
D. -2<x<2
E. -2<x<0

=> -8 < x-3|x| <8

1) x >= 0
-8 < x-3x < 8
-8 < -2x < 8
-4 < x < 4
0 <= x < 4


2) x < 0
-8 < x-3|x| < 8
-8 < x+3x < 8
-8 < 4x < 8
-2 < x < 2
-2 < x < 0

Thus, -2 < x < 4

Ans: C

If 3^x>10, which of the following must be true?

I. x>2
II. x>3
III. x>4

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

=> 3^x > 10 > 3^2
Thus x > 2

Ans: A

If x-7=√x+√7, x=?

A. 8+2√7
B. 8-2√7
C. 8+√7
D. 8-√7
E. 7+2√6

=>x-7=√x+√7
(√x+√7)(√x-√7)=√x+√7
√x-√7 = 1
√x = √7 + 1
x = (√7 + 1)^2
x = 7 + 2√7 + 1 = 8 + 2√7

Ans: A

(1/11)+(1/12)+……+(1/19)+(1/20) is including in which of the following ranges?

A. 0 ~ (1/5)
B. (1/5) ~ (1/4)
C. (1/4)~(1/3)
D. (1/3)~(1/2)
E. (1/2)~1

=> (1/20) < (1/11), (1/20) < (1/12), …, (1/20) < (1/19).
(1/20) + (1/20) + … + (1/20) < (1/11)+(1/12)+……+(1/19)+(1/20)
(1/20)*10 < (1/11)+(1/12)+……+(1/19)+(1/20)
(1/2) < (1/11)+(1/12)+……+(1/19)+(1/20)

(1/11) < (1/10), (1/12) < (1/10), … , (1/20) < (1/10)
(1/11)+(1/12)+……+(1/19)+(1/20) < (1/10) + (1/10) + … + (1/10) = 1

Thus, 1/2 < (1/11)+(1/12)+……+(1/19)+(1/20) < 1

Ans: E

If n is the sum of the first 40 positive integers, what is the greatest prime factor of n?

A. 29
B. 31
C. 37
D. 41
E. 43

=> 1 + 2 + … + 40 = (40*41)/2 = 2^3*5*41
Thus the greatest prime factor of n is 41.

Ans: D

Which of the following is the closest to 11*1010–4*104?

A. 10^2
B. 10^7
C. 10^8
D. 10^9
E. 10^11

=> 4*10^4 is relative small compared with 11*10^10. Thus we can consider 11*10^10 only.
It is close 10^11.

Ans: E

[GMAT math practice question]

What is the unit digit of 3^5+4^5+5^5+6^5?

A. 5
B. 6
C. 7
D. 8
E. 9

=> 3^1 has a unit digit 3.
3^2 has a unit digit 9.
3^3 has a unit digit 7.
3^4 has a unit digit 1.
3^5 has a unit digit 3.

4^1 has a unit digit 4.
4^2 has a unit digit 6.
4^3 has a unit digit 4.
4^4 has a unit digit 6.
4^5 has a unit digit 4.

5^1 has a unit digit 5.
5^2 has a unit digit 5.
5^n has a unit digit 5 for all positive integers n.

6^1 has a unit digit 6.
6^2 has a unit digit 6.
6^n has a unit digit 6 for positive integers n.

Thus 3 + 4 + 5 + 6 = 18 has the unit digit 8.

Ans: D

[GMAT math practice question]

If 3^x-3^{x-2}=648, x=?

A. 5
B. 6
C. 7
D. 8
E. 9

=>

3^x-3^{x-2} = 3^{x-2}3^2-3^{x-2} = 3^{x-2}(3^2-1) = 3^{x-2} x 8 = 648
3^{x-2} = 81 = 3^4
x – 2 = 4
x = 6

Ans: B

[GMAT math practice question]

There are 100 black balls, 100 red balls and 100 green balls evenly mixed in a jar. What is the minimum number of balls to ensure that at least 15 beans selected are the same color?

A. 40
B. 41
C. 42
D. 43
E. 44

=>
We can assume we have 14 black balls, 14 red balls and 14 green balls. After that, if we have just one more ball, it ensures we have 15 balls with a same color.

Thus we need 43 ( = (15-1) + (15-1) + (15-1) + 1 = 14 + 14 + 14 + 1 ) balls.40 balls ensure there are 15 balls that have a same color.

Ans: D

If the gross profit of a certain company is 20 percent of the revenue from sales of the company, what percent of the cost is the gross profit?

A. 20%
B. 25%
C. 30%
D. 35%
E. 40%

=>
Let R be the revenue of the company. The profit is 0.2*R and the cost is 0.8*R.
The answer is (0.2*R) / (0.8*R) * 100 = 0.25 * 100 = 25(%).

Ans: B

[GMAT math practice question]

(7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from 0 to 6, inclusively, b=?

A. 0
B. 1
C. 2
D. 3
E. 4

=>

1045 = 7*149+2 = 7*(7*21 + 2) +2 = 7*(7*(7*3+0) + 2) + 2
= 7*((7^2)*3 +(7^1)*0 + 2) + 2 = (7^3)*3 +(72)*0 + (7^1)*2 + 2

Thus we have b = 0.

Ans: A


[GMAT math practice question]

I. 9/10, 10/11, 11/12, 12/13, 13/14
II. 10/9, 11/10, 12/11, 13/12, 14/13
III. 6/10, 7/10, 8/10, 9/10, 10/10

For which of the following options is the average (arithmetic mean) of the numbers less than the median of numbers?

A. I. only
B. II. only
C. III. only
D. II and III
E. I, II and III

=>

I. Differences between consecutive terms are 10/11 – 9/10 = 1/110, 11/12 – 10/11 = 1/132, 12/13 – 11/12 = 1/156, 13/14 – 12/13 = 1/182. The differences are getting smaller. The average is smaller than the median.

II. Difference between consecutive terms are 11/10 – 10/9 = -1/90, 12/11 – 11/10 = -1/110, 13/12 – 12/11 = -1/132, 14/13 – 13/12 = -1/156. The differences are getting bigger. The average is bigger than the media.

II Difference between consecutive terms are 7/10 – 6/10 = 1/10, 8/10 – 7/10 = 1/10, 9/10 – 8/10 = 1/10, 10/10 – 9/10 = 1/10. All differences are equal. The average is equal to the median.

Ans: A

[GMAT math practice question]

If x-5=√x+√5 , x=?

A. 6+2√5
B. 6-2√5
C. 6+√5
D. 6-√5
E. 5+2√6


=>
x-5=√x+√5
(√x+√5)(√x-√5)=√x+√5
√x-√5 = 1
√x = 1+√5
x = 1 + 2√5 + 5 = 6 + 2√5

Ans: A

Which of the following options is closest to (3^{10})(33^{10})?

A. 10^{10}
B. 10^{14}
C. 10^{18}
D. 10^{20}
E. 10^{22}


=>

(3^{10})(33^{10}) = (3^{10})(3*11)^{10} = (3^{10})(3^{10})(11^{10})= (9^{10})(11^{10}) ~ (10^{10})(10^{10}) = 10^{20}

Ans: D

3 girls and 4 boys line up in a row. If girls stand at the first and boys stand at the last, how many possible cases are there?

A. 96
B. 108
C. 144
D. 192
E. 256

=>

The number of ways that 3 girls permute is 3! = 6 and the number of ways that 4 boys permute is 4! = 24. Those cases happen together and so the number of cases that 3 girls stand at the first and 4 boys stand at the last is 3! * 4! = 6 * 24 = 144.

Ans: C